After reading through all of the above, I don't find a question that needs to be answered.
But I just want to say:
Yep. Uh huh. Fer sher. You are true. Words of higher veracity are unlikely to be found. Every word of that scenario and its description is accurate, and cannot be debated or disputed in any wise.
What's more, I agree, and I thank you for the points.
Answer:
Potential energy increases.
Explanation:
We know that the direction of electric field is from positive to negative charge. As the proton has the positive charge so if it moves in the direction opposite of electric field, it means that the positive charge will move towards the positive region. As the positive charge is attracted towards negative and repelled by the positive charge. So the work done will be negative in bringing the positive charge towards the positive region of the field and potential energy increases in the direction opposite to electric field. As the potential energy decreases in the direction of electric field.
The Lorentz force acting on the proton is:
![F=qvB \sin \theta = qvB](https://tex.z-dn.net/?f=F%3DqvB%20%5Csin%20%5Ctheta%20%3D%20qvB)
(1)
where
q is the proton charge (
![1.6 \cdot 10^{-19} C](https://tex.z-dn.net/?f=1.6%20%5Ccdot%2010%5E%7B-19%7D%20C)
)
v is its speed (7.5 m/s)
B is the magnetic field intensity (1.7 T)
![\theta](https://tex.z-dn.net/?f=%5Ctheta)
is the angle between the direction of v and B, and since the proton is travelling at right angle to the magnetic field,
![\theta=90^{\circ}](https://tex.z-dn.net/?f=%5Ctheta%3D90%5E%7B%5Ccirc%7D)
and
![\sin \theta = 1](https://tex.z-dn.net/?f=%5Csin%20%5Ctheta%20%3D%201)
, so we can ignore it.
The Lorentz force provides the centripetal force that keeps the proton in circular orbit. The centripetal force is
![F_c = m a_c = m \frac{v^2}{r}](https://tex.z-dn.net/?f=F_c%20%3D%20m%20a_c%20%3D%20m%20%20%5Cfrac%7Bv%5E2%7D%7Br%7D%20)
(2)
where
![a_c](https://tex.z-dn.net/?f=a_c)
is the centripetal acceleration
m is the proton mass (
![1.67 \cdot 10^{-27} kg)](https://tex.z-dn.net/?f=1.67%20%5Ccdot%2010%5E%7B-27%7D%20kg%29)
)
v is the proton speed
We said that the Lorentz force provides the centripetal force of the motion, so we can equalize (1) and (2) to find the radius of the proton's orbit:
![m \frac{v^2}{r}=qvB](https://tex.z-dn.net/?f=m%20%5Cfrac%7Bv%5E2%7D%7Br%7D%3DqvB%20)
![r= \frac{mv}{qB}= \frac{(1.67 \cdot 10^{-27} kg)(7.5 m/s)}{(1.6 \cdot 10^{-19}C)(1.7 T)}=4.6 \cdot 10^{-8} m](https://tex.z-dn.net/?f=r%3D%20%5Cfrac%7Bmv%7D%7BqB%7D%3D%20%5Cfrac%7B%281.67%20%5Ccdot%2010%5E%7B-27%7D%20kg%29%287.5%20m%2Fs%29%7D%7B%281.6%20%5Ccdot%2010%5E%7B-19%7DC%29%281.7%20T%29%7D%3D4.6%20%5Ccdot%2010%5E%7B-8%7D%20m%20%20)
Therefore now we can calculate the centripetal acceleration of the proton, which is given by